Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long num = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num * neg;
}
struct node {
  double l, r;
} s[200010];
int n, q;
long long sum[200010];
double getx(int sx, int sy, int tx, int ty) {
  double k = (ty - sy) * 1.0 / (tx - sx);
  return sx - sy * 1.0 / k;
}
inline int findl(double x) {
  int l = 0, r = n, mid;
  while (r > l + 1) {
    int mid = l + r >> 1;
    if (s[mid].r >= x)
      r = mid;
    else
      l = mid;
  }
  return r;
}
inline int findr(double x) {
  int l = 1, r = n + 1, mid;
  while (r > l + 1) {
    int mid = l + r >> 1;
    if (s[mid].l <= x)
      l = mid;
    else
      r = mid;
  }
  return l;
}
int main() {
  int sy = read(), a = read(), b = read();
  n = read();
  for (int i = 1; i <= n; i++) s[i].l = read() * 1.0, s[i].r = read() * 1.0;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + s[i].r - s[i].l;
  q = read();
  while (q--) {
    int x = read(), y = read();
    double l = getx(x, y, a, sy), r = getx(x, y, b, sy), ans;
    int tl = findl(l), tr = findr(r);
    if (s[tl].l > l) l = s[tl].l;
    if (s[tr].r < r) r = s[tr].r;
    if (tl == tr)
      ans = r - l;
    else
      ans = s[tl].r - l + r - s[tr].l + (sum[tr - 1] - sum[tl]) * 1.0;
    ans = ans * 1.0 / y * (y - sy);
    if (ans < 0) ans = 0.0;
    printf("%.10lf\n", ans);
  }
  return 0;
}